Answer:
Routine
Explanation:
Loop Structures — The Method Of Repeating Routines In Statements. Repetition of code are called loops, and they are defined as statements that execute lines of code (or routines) repeatedly according to conditions or iterations. ... Take for example a routine that must write as output the string “Hello” 40 times
Answer:
C. Exist
Hope it helps!
Answer:
The condition is true when their voltage and current specifications with their impedance are matched or complementary to each other.
Explanation:
Solution
Yes it is possible or true to interface an IC with a different technology like the TTL to HCS12 ports. but the condition is that their current and voltage specifications should be matched and their impedance and power also should be matched.
What this implies is that both their voltage and current requirements should be complementary to each other so as their impedance.
Explanation:
First of all get the input from the user, number of rows and number of columns where rows represents seat digit number and column represents the seat letter
rows is initialized to 1 to ensure that row starts at 1 or you can remove it then seat number will start from 0.
The first loop is used for digits starting from 1 to number of rows
The second loop is used for letters starting from 1 to number of columns
since rows and cols are not of the same type that's why we are converting the int type to string type
print(str(rows)+cols) counter will keep updating the columns A, B, C.....
rows= rows + 1 counter will keep updating the rows 1, 2, 3....
Code:
Please refer to the attached image.
Output:
Please enter the number of rows: 2
Please enter the number of columns: 3
1A
1B
1C
2A
2B
2C
Answer:
6.99 x 10⁻³ m³ / s
Explanation:
Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.
h = difference in the height of water column at two ends of delivery pipe
6 - 1 = 5 m
Velocity of flow of water
v = √2gh
= √ (2 x 9.8 x 5)
= 9.9 m /s
Volume of water flowing per unit time
velocity x cross sectional area
= 9.9 x 3.14 x .015²
= 6.99 x 10⁻³ m³ / s